Evaluate $\int^{\pi/2}_0\sin^5x\,dx\,$.
Solution: The integrand has an odd power of $~\sin x\,$. We keep one factor of $~\sin x~$ by itself, and rewrite the remaining $~\sin^4x~$ using the identity $~\sin^2x=1-\cos^2x\,$. $ \int^{\pi/2}_0\sin^5x\,dx= \int^{\pi/2}_0 (\sin^2 x)^2 \cdot \sin x \,dx $ $ = \int^{\pi/2}_0 (1-\cos^2 x)^2 \sin x\, dx$ We can now use a $~u$ -substitution where $ u=\cos x~~~~$ and $~~~~du =-\sin x\,dx\,$. We also need to change our limits of integration. $ x=\dfrac\pi2 \Rightarrow u=0$ $ x=0 \Rightarrow u=1$ Thus $ \int^{\pi/2}_0\sin^5x\,dx=-\int^0_1(1-u^2)^2\,du\,$. We evaluate this definite integral. $ ~-\int^0_1(1-u^2)^2\,du = \int_0^1\big(u^4-2u^2+1\big)\,du$ $ =\left(\dfrac{u^5}5-\dfrac{2u^3}3+u\right)\Bigg|_0^1$ $ =\Big(\dfrac15-\dfrac23+1\Big)=\dfrac8{15}$